Hello, juiicycouture!

I got a different answer for (B) . . .

(B) A merchant ship traveling due north at 12km/h crosses the track of a second ship

travelling due east at 9km/h, at a time 100 min after the second ship passed that point.

Find the distance of the closest approach of the two ships. Code:

B *
| *
| * z
12t | *
| *
| *
* - - - - - - * - - - - - - - - - - *
O 15 P 9t A

Ship A starts at point O and travels east for 100 minutes at 9 km/h.

. . It moves to point P: OP = 15 km.

In the next t hours, it travels 9t km to point A.

Ship B starts at point O travels north at 12 km/h.

In the next t hours, it travels 12t km to point B.

Let z = AB: . z² .= .(9t + 15)² + (12t)² .= .225t² + 270t + 225 .**[1]**

Then: .2z·z' .= .450t + 270 . → . z' .= .(225t + 135)/z .= .0

Hence: .225t + 135 .= .0 . → . t .= -0.6

The ships are closest 0.6 hours (36 minutes) __before__ they crossed paths.

Substitute into [1]: .z² .= .225(-0.6)² + 270(-0.6) + 225 .= .144

. . Therefore: .z .= .12 km.